Title: Daniel's Flashcards / Year 4 / Term 2 / Analytic Number Theory

html: Theorem (Euler). ∑ ⁡ p 1 p = ∞.
proof: Proof. Consider p N = ∏ ⁡ p N ( 1 + 1 p + 1 p 2 + ⋯ + 1 p N ), for N ∈ ℕ = { 1 , 2 , 3 , … }. We have:
p N ≥ ∑ n = 1 N 1 n ≥ ∑ n = 1 N − 1 ∫ n n + 1 d t t = ∫ 1 N − 1 d t t = log ⁡ ( N − 1 )
On the other hand, using 1 + x ≤ e x, so
p N ≤ ∏ p ≤ N exp ⁡ ( 1 p + 1 p 2 + ⋯ + 1 p N ) = exp ⁡ ( ∑ p ≤ N ( 1 p + 1 p 2 + ⋯ + 1 p N ) ) ≤ exp ⁡ ( ∑ p ≤ N ( 1 p + 1 p 2 − p ) ) ≤ exp ⁡ ( C + ∑ p ≤ N 1 p )
Comparing these two bounds gives
∑ p ≤ N 1 p ≥ log ⁡ log ⁡ ( N − 1 ) − C .
Then letting N → ∞ gives the desired result. □
nothing: True

html: Definition 1.1 (Big O and little o notation). Let f , g , h : S → ℂ, S ⊆ ℂ.
Write f ( x ) = O ( g ( x ) ) if there is c > 0 such that | f ( x ) | ≤ c | g ( x ) | for all x ∈ S.
Write f ( x ) = o ( g ( x ) ) if for any 𝜀 > 0 there is x 𝜀 > 0 such that | f ( x ) | ≤ 𝜀 | g ( x ) | for x ∈ S, | x | ≥ x 𝜀.
Write f ( x ) = g ( x ) + O ( h ( x ) ) if f ( x ) − g ( x ) = O ( h ( x ) ) and write f ( x ) = g ( x ) + o ( h ( x ) ) if f ( x ) − g ( x ) = o ( h ( x ) ).
nothing: True

html: Definition 1.2 (Vinogradov notation). Let f , g , h : S → ℂ, S ⊆ ℂ.
Write f ( x ) ≪ g ( x ) or g ( x ) ≫ f ( x ) if f ( x ) = O ( g ( x ) ).
nothing: True

html: Lemma 1.3 (Partial Summation). Assuming that:
- ( a n ) n ∈ ℕ are complex numbers
- x ≥ y ≥ 0
- f : [ y , x ] → ℂ is continuously differentiable
Then
∑ y < n ≤ x a n f ( n ) = A ( x ) f ( x ) − A ( y ) f ( y ) − ∫ y x A ( t ) f ′ ( t ) d t ,
where for t ≥ 1, we define
A ( t ) = ∑ n ≤ t a n = ∑ n = 1 ⌊ t ⌋ a n .
proof: Proof. It suffices to prove the y = 0 case, since then
∑ y < n ≤ x a n f ( n ) = ∑ 0 < n ≤ x a n f ( n ) − ∑ 0 < n ≤ y a n f ( n ) .
Suppose y = 0. By the fundamental theorem of calculus,
𝟙 f ( n ) = f ( x ) − ∫ n x f ′ ( t ) d t = ∫ 0 x f ′ ( t ) 𝟙 [ n , x ] ( t ) d t .
Summing over n ≤ x, we get
𝟙 ∑ n ≤ x a n f ( n ) = A ( x ) f ( x ) − ∫ 0 x f ′ ( t ) ( ∑ n ≤ x 𝟙 [ n , x ] ( t ) a n ) d t = A ( x ) f ( x ) − ∫ 0 x f ′ ( t ) A ( t ) d t □
nothing: True

html: Lemma. If x ≥ 1, then
∑ n ≤ x 1 n = log ⁡ x + γ + O ( 1 x ) ,
where γ ∈ ℝ is Euler’s constant, which is given by γ = lim ⁡ N → ∞ ∑ ⁡ k = 1 N 1 k − log ⁡ ⁡ N.
proof: Proof. Apply Partial Summation with a n = 1, f ( t ) = 1 t, y = 1 2. Clearly A ( t ) = ⌊ t ⌋. Then,
∑ n ≤ x 1 n = ⌊ x ⌋ x + ∫ 1 x ⌊ t ⌋ t 2 d t = x + O ( 1 ) x + ∫ 1 x t − { t } t 2 d t = 1 + O ( 1 x ) + log ⁡ x − ∫ 1 x { t } t 2 d t = 1 + log ⁡ x − ∫ 1 ∞ { t } t 2 d t + O ( 1 x )
The last equality is true since ∫ x ∞ { t } t 2 d t ≤ ∫ x ∞ 1 t 2 d t = 1 x.
Let γ = 1 − ∫ 1 ∞ { t } t 2 d t. Then we have the asymptotic equation as desired.
Taking x → ∞ in the formula, we see that γ is equal to the formula for Euler’s constant, as desired. □
cloze1: True

html: Lemma. For x ≥ 1,
∑ p ≤ x 1 p = ∫ 1 x π ( t ) t 2 d t + O ( 1 ) .
proof: Proof. Apply Lemma 1.3 with 𝟙 a n = 𝟙 ℙ ( n ) (where ℙ is the set of primes), f ( t ) = 1 t, and y = 1.
We get A ( t ) = π ( t ), and then
∑ p ≤ x 1 p = π ( x ) x + ∫ 1 x π ( t ) t 2 d t = ∫ 1 x π ( t ) t 2 d t + O ( 1 ) □
cloze1: True

html: Definition (Arithmetic function). An arithmetic function is a function f : ℕ → ℂ.
nothing: True

html: Definition (Multiplicative). An arithmetic function f is multiplicative if f ( 1 ) = 1 and f ( m n ) = f ( m ) f ( n ) whenever m , n ∈ ℕ are coprime.
Moreover, f is completely multiplicative if f ( m n ) = f ( m ) f ( n ) for all m , n.
nothing: True

html: Definition (Dirichlet convolution). For f , g : ℕ → ℂ, we define
( f ∗ g ) ( n ) = ∑ d | n f ( d ) g ( n d ) ,
where ∑ ⁡ d | n means sum over the divisors of n.
nothing: True

html: Lemma 1.4. The space of arithmetic functions with operations +, ∗ is a commutative ring.
proof: Proof. Since arithmetic functions with + form an abelian group, it suffices to show:
- (i) ( f ∗ g ) ∗ h = f ∗ ( g ∗ h )
- (ii) f ∗ g = g ∗ f
- (iii) f ∗ I = f
- (iv) f ∗ ( g + h ) = f ∗ g + f ∗ h
Proofs:
- (i) Follows from ( f ∗ g ) ( n ) = ∑ ⁡ n = a b f ( a ) g ( b ).
- (ii) Follows from ( f ∗ g ) ( n ) = ∑ ⁡ n = a b f ( a ) g ( b ).
- (iii) Take
I ( n ) = { 1 n = 1 0 n ≠ 0 .
Then one can check that f ∗ I = f.
- (iv) From definition. □
cloze1: True

html: Lemma. The set of arithmetic functions f with f ( 1 ) ≠ 0 form an abelian group with operation ∗.
proof: Proof. Need g such that f ∗ g = I.
( f ∗ g ) ( 1 ) = f ( 1 ) g ( 1 ) = 1 ⟹ g ( 1 ) = 1 f ( 1 ) .
Assume g ( m ) defined for m < n. We will defined g ( n ).
( f ∗ g ) ( n ) = g ( n ) f ( 1 ) + ∑ d | n d ≠ 1 f ( d ) g ( n d ) ⟹ g ( n ) = − 1 f ( 1 ) ∑ d | n d ≠ 1 f ( d ) g ( n d ) ⏟ < n □

html: Lemma. Multiplicative arithmetic functions form an abelian group with ∗. Moreover, for completely multiplicative f, the Dirichlet inverse is μ f.
proof: Proof. For the first part, suffices to show closedness. Let f , g : ℕ → ℂ be multiplicative, and m , n coprime.
If m n = a b, we can write a = a 1 a 2, b = b 1 b 2 where a 1 = ( a , m ), a 2 = ( a , n ), b 1 = ( b , m ) and b 2 = ( b , n ). Therefore we have:
( f ∗ g ) ( m n ) = ∑ m n = a b f ( a ) g ( b ) = ∑ m = a 1 b 1 n = a 2 b 2 f ( a 1 a 2 ) g ( b 1 b 2 ) (above observation) = ∑ m = a 1 b 1 n = a 2 b 2 f ( a 1 ) f ( a 2 ) g ( b 1 ) g ( b 2 ) (by multiplicativity) = ( f ∗ g ) ( m ) ( f ∗ g ) ( n )
(also need to check that inverses are multiplicative).
Now, remains to show that for completely multiplicative f, f ∗ μ f = I.
Note that f ∗ μ f is multiplicative by the first part. So enough to show ( f ∗ μ f ) ( p k ) = I ( p k ) for prime powers p k. Calculate:
f ∗ μ f ( p ) = f ( p ) + μ f ( p ) = f ( p ) − f ( p ) = 0 = I ( p )
and for k ≥ 2:
f ∗ μ f ( p k ) = f ( p k ) + μ f ( p ) f ( p k − 1 ) = f ( p k ) − f ( p ) f ( p k − 1 ) = f ( p ) k − f ( p ) f ( p ) k − 1 = 0 = I ( p k ) □

html: Definition (von Mangoldt function). The von Mangoldt function Λ : ℕ → ℝ is
Λ ( n ) = { log ⁡ p if n = p k for some prime p 0 otherwise
Then log ⁡ = 1 ∗ Λ since
log ⁡ n = log ⁡ ∏ p | n p α p ( n ) = ∑ p | n α p ( n ) log ⁡ p = 1 ∗ Λ ( n ) .
Since μ is the inverse of 1, we have
log ⁡ ∗ μ = 1 ∗ Λ ∗ μ = Λ ∗ 1 ∗ μ = Λ ∗ I = Λ .
nothing: True

html: Definition (Formal series). For f : ℕ → ℂ, define a (formal) series
D f ( s ) = ∑ n = 1 ∞ f ( n ) n − s
for s ∈ ℂ.
nothing: True

html: Lemma. Assuming that:
- f : ℕ → ℂ satisfying | f ( n ) | ≤ n o ( 1 )
Then D f ( s ) converges absolutely for Re ⁡ ( s ) > 1 and defines an analytic function for Re ⁡ ( s ) > 1.
proof: Proof. Let 𝜀 > 0 (fixed), n ∈ ℕ and Re ⁡ ( s ) > 1 + 2 𝜀.
Then
∑ n = N ∞ | f ( n ) n − s | = ∑ n = N ∞ | f ( n ) | n − Re ⁡ ( s ) ≪ ∑ n = N ∞ n 𝜀 − Re ⁡ ( s ) ≤ ∑ n = N ∞ n − 1 − 𝜀 ≤ N − 1 − 𝜀 + ∑ n = N + 1 ∞ ∫ n − 1 n t − 1 − 𝜀 d t ≤ N − 1 − 𝜀 + ∫ N ∞ t − 1 − 𝜀 d t ≪ N − 𝜀
Hence we have absolute convergence for Re ⁡ ( s ) > 1 + 2 𝜀. Also, D f ( s ) is a uniform limit of the functions ∑ ⁡ n = 1 N f ( n ) n − s. From complex analysis, a uniform limit of analytic functions is analytic. Hence D f ( s ) is analytic for Re ⁡ ( s ) > 1 + 2 𝜀.
Now let 𝜀 → 0. □
cloze1: True

html: Theorem (Euler product). Assuming that:
- f : ℕ → ℂ be bounded (| f ( n ) | ≪ 1) and multiplicative
Then
D f ( s ) = ∏ p ( 1 + f ( p ) p 2 + f ( p 2 ) p 2 s + ⋯ ) ,
for Re ⁡ ( s ) > 1. Furthermore, if f is completely multiplicative, then
D f ( s ) = ∏ p ( 1 − f ( p ) p s ) − 1 ,
for Re ⁡ ( s ) > 1.
proof: Proof. Let N ∈ ℕ, Re ⁡ ( s ) = σ > 1. Let
D f ( s , N ) = ∏ p ≤ N ( 1 + f ( p ) p s + f ( p 2 ) p 2 s + ⋯ ) .
Note that the series defining the factors are absolutely convergent, since
∑ k = 1 ∞ | f ( p k ) | | p k s | ≪ ∑ k = 1 ∞ 1 p k s < ∞
(geometric series).
Therefore, multiplying out,
D f ( s , N ) = ∑ n = 1 ∞ a ( n , N ) f ( n ) n − s
where
a ( n , N ) = # { ways to write n as a product of prime powers, where the primes are ≤ N } .
The fundamental theorem of arithmetic tells us that a ( n , N ) ∈ { 0 , 1 } and a ( n , N ) = 1 for n ≤ N.
Now,
| D f ( S ) − D f ( s , N ) | ≤ ∑ n = N + 1 ∞ | f ( n ) | | n − s | ≪ ∑ n = N + 1 ∞ n − σ → N → ∞ 0
(since ∑ ⁡ n = 1 ∞ n − σ < ∞). Hence D f ( s ) = lim ⁡ N → ∞ D f ( s , N ).
Finally, for f completely multiplicative, use geometric formula:
∑ k = 1 ∞ f ( p ) k p k s = 1 1 − f ( p ) p s . □
nothing: True

html: Lemma. Assuming that:
- f , g : ℕ → ℂ
- | f ( n ) | , | g ( n ) | ≤ n o ( 1 )
Then
D f ∗ g ( s ) = D f ( s ) D g ( s )
for Re ⁡ ( s ) > 1.
proof: Proof. We know D f ( s ) and D g ( s ) are absolutely convergent, so can expand out the product.
D f ( s ) D g ( s ) = ∑ a , b = 1 ∞ f ( a ) g ( b ) ( a b ) − s = ∑ n = 1 ∞ ∑ n = a b f ( a ) g ( b ) n − s = ∑ n = 1 ∞ ( f ∗ g ) ( n ) n − s = D f ∗ g ( s ) □
cloze1: True

html: Definition (Riemann zeta function). For Re ⁡ ( s ) > 1, define
ζ ( s ) = ∑ n = 1 ∞ n − s .
nothing: True

html: Lemma (Dirichlet hyperbola method). Assuming that:
- f , g : ℕ → ℂ
- x ≥ y ≥ 1
Then
∑ n ≤ x ( f ∗ g ) ( n ) = ∑ d ≤ y f ( d ) ∑ m ≤ x d g ( m ) + ∑ m ≤ x y g ( m ) ∑ y < d ≤ x m f ( d ) .
proof: Proof. ∑ ⁡ n ≤ x ( f ∗ g ) ( n ) = ∑ d m ≤ x f ( d ) g ( m ). Split this sum into parts with d ≤ y and d > y to get the conclusion. □
nothing: True

html: Theorem 1.5 (Dirichlet’s divisor problem). For x ≥ 2,
∑ n ≤ x τ ( n ) = x log ⁡ x + ( 2 γ − 1 ) x + O ( x 1 ∕ 2 )
where γ is Euler’s constant.
proof: Proof. We use the Dirichlet hyperbola method, with y = x 1 2. Note that τ = 1 ∗ 1. Then,
∑ n ≤ x τ ( n ) = ∑ d ≤ x 1 2 ∑ m ≤ x d 1 + ∑ m ≤ x 1 2 ∑ x 1 2 < d ≤ x m 1 = ∑ x ≤ x 1 2 ( x d + O ( 1 ) ) + ∑ m ≤ x 1 2 ( x m − x 1 2 + O ( 1 ) ) = x ∑ d ≤ x 1 2 1 d + x ∑ m ≤ x 1 2 1 m − ∑ m ≤ x 1 2 x 1 2 + O ( x 1 2 )
Recall from Lecture 2 that
∑ d ≤ y 1 d = log ⁡ y + γ + O ( 1 y ) .
Taking y = x 1 2, the previous expression becomes
2 × ( 1 2 log ⁡ x + γ + O ( 1 x 1 2 ) ) − x + O ( x 1 2 ) = x log ⁡ x + ( 2 γ − 1 ) x + O ( x 1 2 ) . □
nothing: True

html: Theorem 2.1 (Merten’s Theorem). Assuming that:
- x ≥ 3
Then
- (i) ∑ ⁡ p ≤ x log ⁡ p p = log ⁡ x + O ( 1 )
- (ii) ∑ ⁡ p ≤ x 1 p = log ⁡ log ⁡ x + M + O ( 1 log ⁡ x ) (for some M ∈ ℝ)
- (iii) ∏ ⁡ p ≤ x ( 1 − 1 p ) = c + o ( 1 ) log ⁡ x (for some c > 0)
proof: Proof.
- (i) Let N = ⌊ x ⌋. Consider N !. We have N N ≤ N N N ! ≥ N N e − N
(the second inequality cn be proved by induction, using ( 1 + 1 N ) N ≤ e).
Let v p ( k ) be the largest power of p dividing k. Then
N ! = ∏ p ≤ N p v p ( N ! ) (fundamental theorem of arithmetic) ⟹ log ⁡ N ! = ∑ p ≤ N v p ( N ! ) log ⁡ p
We have v p ( N ! ) = ∑ ⁡ j = 1 ∞ ⌊ N p j ⌋. Indeed,
𝟙 𝟙 v p ( N ! ) = ∑ k = 1 N v p ( k ) = ∑ k = 1 N ∑ j = 1 ∞ 𝟙 v p ( k ) ≥ j = ∑ j = 1 ∞ ∑ k = 1 N 𝟙 v p ( k ) ≥ j = ∑ j = 1 ∞ ⌊ N p j ⌋
Now,
log ⁡ N ! = ∑ p ≤ N ⌊ N p ⌋ log ⁡ p + ∑ p ≤ N ( log ⁡ p ) ∑ j = 2 ∞ ⌊ N p j ⌋ = ∑ p ≤ N ( N p + O ( 1 ) ) log ⁡ p + O ( ∑ p ≤ N log ⁡ p p 2 ⋅ N ) (geometric series) = N ∑ p ≤ N log ⁡ p p + O ( ∑ p ≤ N log ⁡ p ) ⏟ ≤ ( log ⁡ ⁡ N ) π ( N ) ≪ N ( C h e b y s h e v ) + O ( N ) (since ∑ p log ⁡ p p 2 < ∞ ) = N ∑ p ≤ N log ⁡ p p + O ( N )
Combine with log ⁡ N ! = N log ⁡ N + O ( N ) to get
N log ⁡ N + O ( N ) = N ∑ p ≤ N log ⁡ p p + O ( N ) .
Divide by N to get the result.
- (ii) By Partial Summation, this is
1 + O ( 1 log ⁡ x ) + ∫ 2 x ∑ p ≤ t log ⁡ p p t ( log ⁡ t ) 2 d t .
Writing 𝜀 ( t ) = ∑ ⁡ p ≤ t log ⁡ p p − log ⁡ t, we get
1 + O ( 1 log ⁡ x ) + ∫ 2 x d t t log ⁡ t + ∫ 2 x 𝜀 ( t ) t ( log ⁡ t ) 2 d t = 1 + O ( 1 log ⁡ x ) + log ⁡ log ⁡ x − log ⁡ log ⁡ 2 + ∫ 2 ∞ 𝜀 ( t ) t ( log ⁡ t ) 2 d t + O ( ∫ x ∞ | 𝜀 ( t ) | t ( log ⁡ t ) 2 d t )
By part (i),
∫ x ∞ | 𝜀 ( t ) | t ( log ⁡ t ) 2 d t ≪ ∫ x ∞ 1 t ( log ⁡ t ) 2 d t ≪ 1 log ⁡ x
Take M = 1 − log ⁡ log ⁡ 2 + ∫ 2 ∞ 𝜀 ( t ) t ( log ⁡ t ) 2 d t.
- (iii) Use Taylor expansion
log ⁡ ( 1 − y ) = − ∑ j = 1 ∞ y j j
to get
log ⁡ ∏ p ≤ x ( 1 − 1 p ) = ∑ p ≤ x log ⁡ ( 1 − 1 p ) = − ∑ p ≤ x 1 p − ∑ p ≤ x ∑ j = 2 ∞ p − j j
Write H = ∑ ⁡ p ∑ j = 2 ∞ p − j j. We get
= − ∑ p ≤ x 1 p − H + O ( ∑ p > x ∑ j = 2 ∞ p − j j ⏟ ≪ p − 2 ) = − ∑ p ≤ x 1 p − H + O ( 1 x ) = (ii) − log ⁡ log ⁡ x − H + O ( 1 x )
Taking exponentials,
∏ p ≤ x ( 1 − 1 p ) = c + o ( 1 ) log ⁡ x . □
nothing: True

html: Definition (Sieve problem). Let P ⊆ ℙ. Let
P ( z ) = ∏ p ∈ P p ≤ z p ,
and let A ⊆ ℤ. Denote
S ( A , P , z ) = | { n ∈ A : ( n , P ( z ) ) = 1 } | .
nothing: True

html: Lemma. We have
S ( A , P , z ) = ∑ d | P ( z ) μ ( d ) | A d | .
proof: Proof. Recall that
𝟙 𝟙 n = 1 = ( μ ∗ 1 ) ( n )
(since μ is the inverse of 1). Hence,
𝟙 S ( A , P , z ) = ∑ n ∈ A 𝟙 ( n , P ( z ) ) = 1 = ∑ n ∈ A ∑ d | P ( z ) d | n μ ( d ) = ∑ d | P ( z ) μ ( a ) | A d | □

html: Theorem (Sieve of Erastothenes – Legendre). Assuming that:
- A ⊆ [ 1 , x ] ∩ ℕ
- 2 ≤ z ≤ x
- Assume the Sieve Hypothesis
Then S ( A , P , z ) = | A | ∏ p ≤ 2 p ∈ P ( 1 − g ( p ) ) + O ( x 1 2 ( log ⁡ x ) 1 2 2 − log ⁡ x 4 log ⁡ z ( ∑ d ≤ x d | P ( z ) | R d | 2 ) 1 2 + | A | e log ⁡ x log ⁡ z ∏ p ≤ z p ∈ P ( 1 + g ( p ) ) e )
proof: Proof. Recall from previous lecture that
S ( A , P , z ) = ∑ d | P ( z ) d ≤ x μ ( d ) | A d | (since A d = ∅ for d > x ) = ∑ d | P ( z ) d ≤ x μ ( d ) g ( d ) | A | + O ( ∑ d ≤ x d | P ( z ) | R d | ) (sieve hypothesis) = | A | ∑ d | P ( z ) μ ( d ) g ( d ) + O ( ∑ d ≤ x d | P ( z ) | R d | + | A | ∑ d | P ( z ) d > x g ( d ) ) = | A | ∏ p ≤ z p ∈ P ( 1 − g ( p ) ) + O ( ∑ d ≤ x d | P ( z ) | R d | + | A | ∑ d | P ( z ) d > x g ( d ) )
We estimate the first error term using Cauchy-Schwarz:
∑ d ≤ x d | P ( z ) | R d | ≤ ( ∑ d ≤ x d | P ( z ) | R d | 2 ) 1 2 ( ∑ d ≤ x d | P ( z ) 1 ) 1 2 .
Note that if d | P ( z ), d > x 1 2, then
ω ( d ) ≥ log ⁡ x 1 2 log ⁡ z ,
(ω ( d ) – number of distinct prime factors of d), since z ω ( d ) ≥ d ≥ x 1 2.
Hence, 2 ω ( d ) ≥ 2 log ⁡ x 2 log ⁡ z, d | P ( z ), d > x 1 2.
Now we get
∑ d ≤ x d | P ( z ) 1 ≤ 2 − log ⁡ x 2 log ⁡ z ∑ d ≤ x d | P ( z ) 2 ω ( d ) ⏟ = τ ( d ) ≤ 2 − log ⁡ x 2 log ⁡ z ∑ d ≤ x τ ( d ) ≪ 2 − log ⁡ x 2 log ⁡ z x log ⁡ x
(the last step is by Dirichlet’s divisor problem: ∑ ⁡ d ≤ x τ ( d ) = ( 1 + o ( 1 ) ) ⋅ x log ⁡ x).
Substituting this in, the first error term becomes as desired.
Now we estimate the second error term. We have
∑ d | P ( z ) d > x g ( d ) ≤ x − 1 log ⁡ z ∑ d | P ( z ) g ( d ) d 1 log ⁡ z (since d > x in the sum: Rankin’s trick) = x − 1 log ⁡ z ∏ p ≤ z p ∈ P ( 1 + g ( p ) p 1 log ⁡ ⁡ z ⏟ ≤ z 1 log ⁡ ⁡ z = e ) ≤ x − 1 log ⁡ z ∏ p ≤ z p ∈ P ( 1 + e g ( p ) ) ≤ x − 1 log ⁡ ⁡ z ⏟ = e − log ⁡ ⁡ x log ⁡ ⁡ z ∏ p ≤ z p ∈ P ( 1 + g ( p ) ) e ( 1 + e y ≤ ( 1 + y ) e )
Combining the error terms, the claim follows. □
nothing: True

html: Theorem 2.2 (Selberg sieve). Assuming that:
- z ≥ 2
- A ⊆ ℤ finite
- P ⊆ ℙ
- Assume the sieve hypothesis
- h : ℕ → [ 0 , ∞ ) be the multiplicative function supported on square-free numbers, given on the primes by
h ( p ) = { g ( p ) 1 − g ( p ) p ∈ P 0 p ∉ P
Then
S ( A , P , z ) ≤ | A | ∑ d ≤ z h ( d ) + ∑ d ≤ z 2 d | P ( z ) τ 3 ( d ) | R d | .
proof: Proof. Let ( ρ d ) d ∈ ℕ be real numbers with
ρ 1 = 1 , ρ d = 0 , d > z (∗)
Then,
𝟙 𝟙 ( n , P ( z ) ) = 1 ≤ ( ∑ d | n d | P ( z ) ρ d ) 2 .
(If ( n , P ( z ) ) = 1, get 1 ≤ ρ otherwise use 0 ≤ x 2).
Summing over n ∈ A,
𝟙 S ( A , P , z ) = ∑ n ∈ A 𝟙 ( n , P ( z ) ) = 1 ≤ ∑ n ∈ A ( ∑ d | n d | P ( z ) ρ d ) 2 = ∑ d 1 d 2 | P ( z ) ρ d 1 ρ d 2 ∑ n ∈ A [ d 1 , d 2 ] | n 1 = | A | ∑ d 1 , d 2 | P ( z ) ρ d 1 ρ d 2 g ( [ d 1 , d 2 ] ) + ∑ d 1 , d 2 | P ( z ) ρ d 1 ρ d 2 R [ d 1 , d 2 ] ⏟ E (sieve hypothesis)
([ m , n ] means lcm ⁡ ( m , n )).
We first estimate E:
E ≤ max ⁡ k | ρ k | 2 ∑ d 1 , d 2 | P ( z ) | R [ d 1 , d 2 ] | = max ⁡ k | ρ k | 2 ∑ d ≤ z k d | P ( z ) ∑ d 1 , d 2 ∈ ℕ [ d 1 , d 2 ] = d | R d | ( d = [ d 1 , d 2 ] )
We have
∑ d 1 , d 2 ∈ ℕ d = [ d 1 , d 2 ] 1 = ∑ l ≤ z ∑ d 1 ′ , d 2 ′ ∈ ℕ d = l d 1 ′ d 2 ′ ( d 1 ′ , d 2 ′ ) = 1 1 ( l = ( d 1 , d 2 ) , d 1 ′ = d 1 ∕ l , [ d 1 , d 2 ] = l d 1 ′ d 2 ′ ) ≤ τ 3 ( d )
Therefore
E ≤ ∑ d ≤ z z d | P ( z ) τ 3 ( d ) | R d | ⋅ max ⁡ k | ρ k | 2 .
Now it suffices to prove that there is a choice of ( ρ d ) d ∈ ℕ satisfying (∗) such that
Claim 1: ∑ ⁡ d 1 , d 2 | P ( z ) ρ d 1 ρ d 2 g ( [ d 1 , d 2 ] ) = 1 ∑ d ≤ z h ( d ).
Claim 2: | ρ k | ≤ 1 for all k ∈ ℕ.
Proof of claim 1:
We have, writing k = ( d 1 , d 2 ), d i ′ = d i k,
∑ d 1 , d 2 | P ( z ) ρ d 1 ρ d 2 g ( [ d 1 , d 2 ] ) = ∑ k | P ( z ) μ ( k ) 2 ∑ d 1 ′ , d 2 ′ | P ( z ) k ( d 1 ′ , d 2 ′ ) = 1 ρ k d 1 ′ ρ k d 2 ′ g ( k d 1 ′ d 2 ′ ) = ∑ k | P ( z ) μ ( k ) 2 g ( k ) ∑ d 1 ′ , d 2 ′ | P ( z ) k ( d 1 ′ , d 2 ′ ) = 1 ρ k d 1 ′ ρ k d 2 ′ g ( d 1 ′ ) g ( d 2 ′ ) (multiplicativity)
We have
𝟙 𝟙 ( d 1 ′ , d 2 ′ ) = 1 = ∑ c | d 1 ′ c | d 2 ′ μ ( c )
(since μ ∗ 1 = I) so the previous expression becomes
∑ k | P ( z ) μ ( k ) 2 g ( k ) ∑ c | P ( z ) k μ ( c ) ∑ d 1 ′ , d 2 ′ | P ( z ) d 1 ′ ≡ 0 ( mod ( ) c ) d 2 ′ ≡ 0 ( mod c ) ρ k d 1 ′ ρ k d 2 ′ g ( d 1 ′ ) g ( d 2 ′ ) = ∑ k | P ( z ) μ ( k ) 2 g ( k ) ∑ c | P ( z ) k μ ( k ) ( ∑ d | P ( z ) k d ≡ 0 ( mod c ) ) 2 = ∑ k | P ( z ) μ ( k ) 2 g ( k ) − 1 ∑ c | P ( z ) k μ ( c ) ( ∑ d ′ | P ( z ) c k ρ c k d ′ g ( c k d ′ ) ) 2 = ∑ n ≤ z h ( m ) − 1 ( ∑ d | P ( z ) d ≡ 0 ( mod n ) ρ d g ( d ) ) 2
(multiplicativity, d = c d ′, m = c k, d = m d ′) since 1 h = μ 2 g ∗ μ (check on primes 1 h ( p ) = 1 g ( p ) − 1).
Now we get
∑ d 1 , d 2 | P ( z ) ρ d 1 ρ d 2 g ( [ d 1 , d 2 ] ) = ∑ m ≤ z h ( m ) − 1 ζ m 2
where
ζ m = ∑ d | P ( z ) d ≡ 0 ( mod m ) ρ d g ( d ) .
Want to minimise this subject to (∗).
We need to translate the condition ρ 1 = 1. Note that
𝟙 ∑ m ≤ z m ≡ 0 ( mod c ) μ ( m ) ζ m = ∑ d | P ( z ) ρ d g ( d ) ∑ m | d c | m μ ( m ) ⏟ ∑ m ′ | d c μ ( c ) μ ( m ′ ) = μ ( d ) 𝟙 d = e = μ ( e ) ρ e g ( e )
Hence,
ρ e = μ ( e ) g ( e ) ∑ m ≤ 2 m ≡ 0 ( mod c ) μ ( m ) ζ m .
Now,
ρ 1 = 1 = ∑ m ≤ z μ ( m ) ζ m .
By Cauchy-Schwarz, we then get
( ∑ m ≤ z h ( m ) − 1 ζ m 2 ) ( ∑ m ≤ z μ ( m ) 2 h ( m ) ) ≥ ( ∑ m ≤ z μ ( m ) ζ m ) 2 = 1 .
Hence
∑ m ≤ z h ( m ) − 1 ζ m 2 ≥ 1 ∑ m ≤ z μ ( m ) 2 h ( m ) = 1 ∑ m ≤ z h ( m ) .
Equality holds for T m = h ( m ) G ( z ), where G ( z ) = ∑ ⁡ m ≤ z h ( m ). We now check that with these ζ m, ρ d = 0 for d > z.
Note that
ρ c = μ ( c ) g ( c ) G ( z ) ∑ m ≤ z m ≡ ( mod c ) μ ( m ) h ( m ) .
Hence, ρ c = 0 for c > 2.
This proves Claim 1.
Now we prove Claim 2 (| ρ c | ≤ 1).
Note that any m has at most one representation as m = e m ′, where e | d, ( m ′ , d ) = 1 (for any d ∈ ℕ).
Now,
G ( z ) ≥ ∑ c | d ∑ m ′ ≤ z c ( m ′ , d ) = 1 h ( c m ′ ) = ∑ c | d h ( c ) ∑ m ′ ≤ z c ( m ′ , d ) = 1 h ( m ′ ) ≥ ∑ c | d h ( e ) ∑ m ′ ≤ z d ( m ′ , d ) = 1 h ( m ′ )
Now,
ρ d = μ ( d ) 2 h ( d ) g ( d ) G ( z ) ∑ m ′ ≤ z d ( m ′ , d ) = 1 μ ( m ′ ) h ( m ′ ) .
Substituting the lower bound for G ( z ),
| ρ d | ≤ h ( d ) g ( d ) ∑ c | d h ( e ) = 1 ,
since 1 ∗ h = h g. □
nothing: True

html: Lemma 2.3. Assuming that:
- z ≥ 3
- g : ℕ → [ 0 , 1 ] multiplicative
- for some K , A ∈ ℝ we have
∑ p ≤ z g ( p ) log ⁡ p ≤ κ log ⁡ z + A .
Then
1 ∑ m ≤ z h ( m ) ≤ 2 ∏ p ≤ z 1 ∕ ( e κ + 1 ) ( 1 − g ( p ) ) ,
where h is defined in terms of g as in Selberg’s sieve.
proof: Proof. Note that for any c ∈ ( 0 , 1 ),
∑ m ≤ z h ( m ) ≥ ∑ m ≤ z m | P ( z c ) h ( m ) = G ( z , c ) .
Then
∏ p ≤ z c p ∈ P ( 1 − g ( p ) ) − 1 − G ( z , c ) = ∏ p ≤ z c p ∈ P ( 1 + h ( p ) ) − ∑ m ≤ z m | P ( z c ) h ( m ) = ∑ m > z m | P ( z c ) h ( m )
By Rankin’s trick,
1 − ∏ p ≤ z c ( 1 − g ( p ) ) G ( z , c ) = ∏ p ≤ z c ( 1 − g ( p ) ) ∑ m > z m | P ( z c ) h ( m ) ≤ ∏ p ≤ z c ( 1 − g ( p ) ) z − λ log ⁡ z ∑ m | P ( z c ) h ( m ) m λ log ⁡ z (for any λ > 0 ) = ∏ p ≤ z c ( 1 − g ( p ) ) e − λ ∏ p ≤ z c ( 1 + h ( p ) p λ log ⁡ z ) = e − λ ∏ p ≤ z c ( 1 − g ( p ) + g ( p ) p λ log ⁡ z ≤ e − λ exp ⁡ ( ∑ p ≤ z c ( p λ log ⁡ ⁡ z − 1 ) ⏟ ≤ λ log ⁡ ⁡ z ( log ⁡ ⁡ p ) p λ log ⁡ ⁡ z ) ≤ exp ⁡ ( − λ + λ log ⁡ ⁡ z ∑ p ≤ z c g ( p ) ( log ⁡ p ) p λ log ⁡ z ) ( 1 + t ≤ e t ) ≤ exp ⁡ ( − λ + c e c λ λ κ + λ e c λ A log ⁡ ⁡ z )
Choose c = 1 λ and λ = e κ + 1 to get the claim. □
cloze1: True

html: Theorem 2.4 (Brun-Titchmarsh Theorem). Assuming that:
- x ≥ 0, y ≥ 2
- 𝜀 > 0 and y is large in terms of 𝜀
Then
π ( x + y ) − π ( x ) ≤ ( 2 + 𝜀 ) y log ⁡ y .
proof: Proof. Apply the Selberg sieve with A = [ x , x + y ] ∩ ℕ, P = ℙ. Note that for any d ≥ 1,
| { a ∈ A : a ≡ 0 ( m o d d ) } = y d + O ( 1 ) .
Hence, the sieve hypothesis holds with g ( d ) = 1 d, R d = O ( 1 ).
Now, the function h in Selberg sieve is given on primes by
h ( p ) = g ( p ) 1 − g ( p ) = 1 p − 1 = 1 φ ( p ) ,
where φ is the Euler totient function. In general,
h ( d ) = μ ( d ) 2 ⋅ 1 φ ( d )
(since φ is multiplicative). Now, for any z ≥ 2, Selberg sieve yields
S ( A , ℙ , z ) ≤ y ∑ d ≤ z μ ( d ) 2 φ ( d ) + O ( ∑ d ≤ z 2 τ 3 ( d ) ) .
By Problem 11 on Example Sheet 1, the error term is O ( z 2 ( log ⁡ z ) 2 ).
Take z = y 1 2 − 𝜀 1 0. Then
z 2 ( log ⁡ z ) 2 ≪ ( y 1 2 − 𝜀 1 0 ) 2 ( log ⁡ y ) 2 ≪ y 1 − 𝜀 2 0
(for y ≥ y 0 ( 𝜀 )). We estimate
∑ d ≤ z μ ( d ) 2 φ ( d ) = ∑ d ≤ z μ ( d ) 2 d d φ ( d ) = ∑ d ≤ z μ ( d ) 2 d ⋅ ∏ p | d ( 1 + 1 p + 1 p 2 + ⋯ ) ⏟ = p p − 1 = p φ ( p ) ≥ ∑ n ≤ z 1 n
since any n ≤ z has at least one representation as
n = d p 1 a 1 ⋯ p k a k ,
where d ≤ z is square-free and p i | d are primes and a 1 ≥ 0.
We have proved
∑ n ≤ z 1 n = log ⁡ z + O ( 1 ) ≥ ( 1 − 1 1 0 ) log ⁡ z
for z ≥ z 0 ( 𝜀 ).
Putting everything together gives us
π ( x + y ) − π ( x ) ≤ S ( A , ℙ , z ) + z ≤ y ( 1 − 𝜀 1 0 ) log ⁡ z + z + y 1 − 𝜀 2 0 ≤ ( 2 + 𝜀 ) y log ⁡ y
for y ≥ y 0 ( 𝜀 ). □
nothing: True

html: Lemma 3.1. Assuming that:
- σ > 1
- t ∈ ℝ
Then | ζ ( σ + i t ) | ≍ σ 1.
proof: Proof. By the Euler product formula,
ζ ( σ + i t ) = ∏ p ( 1 − p − σ − i t ) − 1 ,
hence
| ζ ( σ + i t ) | = ∏ p | 1 − p − σ − i t | − 1 .
By the triangle inequality,
1 − p − σ ≤ | 1 − p − σ − i t | ≤ 1 + | p − σ − i t | = 1 + p − σ .
Hence,
∏ p ( 1 − p − σ ) − 1 ≥ | ζ ( σ + i t ) | ≥ ∏ p ( 1 = p − σ ) − 1 .
Note that these products converge if and only if ∑ ⁡ p p − σ converges, and this sum converges by the comparison test. □
cloze1: True

html: Lemma 3.2 (Polynomial growth of e t a in half-planes).
- (i)
f ( ζ ) extends to a meromorphic function on ℂ, with the only pole being a simple pole which is at s = 1.
- (ii)
Let k ≥ 0 be an integer. Then, for Re ⁡ ( s ) ≥ − k and | s − 1 | ≥ 1 1 0 we have | ζ ( s ) | ≪ k | s | k + 2 + 1.
proof: Proof. Let k ≥ 0 be an integer.
We claim that there exist polynomials P k, Q k of degree ≤ k + 1 and such that for Re ⁡ ( s ) > 1,
ζ ( s ) = 1 s − 1 + Q k ( s ) + s ( s + 1 ) ⋯ ( s + k ) ∫ 1 ∞ P k ( { t } ) t s + k + 1 d t . (∗)
First assume (∗) holds.
Then, since P k ( { t } ) ≪ k 1, for Re ⁡ ( s ) > − k − 1 2, | s − 1 | ≥ 1 1 0, (∗) gives | ζ ( s ) | ≪ k | s | k + 2 + 1. So (ii) follows.
For (i), using analytic continuation and (∗), it suffices to show that the RHS of (∗) is meromorphic for Re ⁡ ( s ) > − k, with the only pole a simple one at s = 1.
Suffices to show that
∫ 1 ∞ P k ( { t } ) t s + k + 1 d t
is analytic for Re ⁡ ( s ) > − k.
This follows from the following criterion: If U ⊆ ℂ is open, f : U × ℝ → ℂ is piecewise continuous, and if s ↦ f ( s , t ) is analytic in U for any t ∈ ℝ, then ∫ ℝ f ( s , t ) d t is analytic in U, provided that ∫ ℝ | f ( s , t ) | d t is bounded on compact subsets of U.
Applying this with 𝟙 f ( s , t ) = P k ( { t } ) t s + k + 1 𝟙 [ 1 , ∞ ) concludes the proof of (i) assuming (∗).
We are left with proving (∗). We use induction on k.
Case k = 0: By Partial Summation,
ζ ( s ) = ∑ n = 1 ∞ 1 n s = s ∫ 1 ∞ ⌊ u ⌋ u s + 1 d u (apply partial summation to ∑ n ≤ x n − s and let x → ∞ ) = s ∫ 1 ∞ 1 u s d u − ∫ 1 ∞ { u } u s + 1 d u = 1 s − 1 + 1 − s ∫ 1 ∞ { u } u s + 1 d u
Take P 0 ( u ) = u, Q 0 ( u ) = 1.
Case k + 1 assuming k: Let
c k = ∫ 0 1 P k ( { t } ) d t .
For Re ⁡ ( s ) > 1, we have
ζ ( s ) = 1 s − 1 + Q k ( s ) + c k s ( s + 1 ) ⋯ ( s + k − 1 ) + s ( s + 1 ) ⋯ ( s + k ) ∫ 1 ∞ P k ( { t } ) − c k t s + k + 1 d t .
Let
P k + 1 ( u ) = − ∫ 0 u ( P k ( t ) − c k ) d t .
This is a polynomial of degree ≤ k + 2. By integration by parts,
∫ 1 ∞ P t ( { t } ) − c k t s + k + 1 d t = ( s + k + 1 ) ∫ 1 ∞ P k + 1 ( { u } ) u s + k + 1 d u .
Substituting this into the previous equation, we get that case k + 1 follows. □
nothing: True

html: Definition (Gamma function). For Re ⁡ ( s ) > 0, let
Γ ( s ) = ∫ 0 ∞ t s − 1 e − t d t .
nothing: True

html: Theorem 3.4 (Functional equation for z e t a). Assuming that:
- ξ ( s ) = 1 2 s ( s − 1 ) π − s ∕ 2 Γ ( s s ) ζ ( s ) for s ∈ ℂ
Then ξ is an entire function and ξ ( s ) = ξ ( 1 − s ) for s ∈ ℂ. Hence
π − s ∕ 2 Γ ( s 2 ) ζ ( s ) = π − 1 − s 2 Γ ( 1 − s 2 ) ζ ( 1 − s )
for s ∈ ℂ ∖ { 0 , 1 }.
proof: Proof. Let Re ⁡ ( s ) > 1. Then,
Γ ( s 2 ) = ∫ 0 ∞ t s 2 − 1 e − t d t .
Make the change of variables f = π n 2 u to get
Γ ( s 2 ) = π s ∕ 2 n s ∫ 0 ∞ u s 2 − 1 e − π n 2 u d u .
Hence
π − s 2 n − s Γ ( s 2 ) = ∫ 0 ∞ u s 2 − 1 e − π n 2 u d u .
Summing over n ∈ ℕ and using Fubini,
π − s 2 Γ ( s 2 ) ζ ( s ) = ∑ n = 1 ∞ ∫ 0 ∞ u s 2 − 1 e − π n 2 u d u = 1 2 ∫ 0 ∞ u s 2 − 1 ( 𝜃 ( u ) − 1 ) d u 𝜃 ( u ) = ∑ n = − ∞ ∞ e − π n 2 u = 1 2 ∫ 0 1 u s 2 − 1 ( 𝜃 ( u ) − 1 ) d u + 1 2 ∫ 1 ∞ u s 2 − 1 ( 𝜃 ( u ) − 1 ) d u ( ∗ )
By the functional equation
𝜃 ( u ) = 1 u 𝜃 ( 1 u ) ,
we have
∫ 0 1 u s 2 − 1 ( 𝜃 ( u ) − 1 ) d u = ∫ 0 1 u s 2 − 3 2 𝜃 ( 1 u ) d u − ∫ 0 1 u s 2 − 1 d u = ∫ 1 ∞ v − s + 1 2 𝜃 ( v ) d v − ∫ 0 1 u s 2 − 1 d u ( u = 1 v ) = ∫ 1 ∞ v − s + 1 2 ( 𝜃 ( v ) − 1 ) d v + ∫ 1 ∞ v − s + 1 2 d v ⏟ 1 s − 1 2 − ∫ 0 1 u s 2 − 1 d u ⏟ 1 s 2
Plugging this into (∗), we get
π − s 2 Γ ( s 2 ) ζ ( s ) = 1 2 ∫ 1 ∞ ( u − s + 1 2 + u s 2 − 1 ) ( 𝜃 ( u ) − 1 ) d u − 1 s ( s − 1 ) .
Hence
ξ ( s ) = − 1 2 + 1 4 s ( s − 1 ) ∫ 1 ∞ ( u − s + 1 2 + u s 2 − 1 ) ( 𝜃 ( u ) − 1 ) d u . (∗∗)
Since | 𝜃 ( u ) − 1 | ≪ e − π u, applying the criterion for integrals of analytic functions being analytic, we see that ξ ( s ) is entire. So by analytic continuation, (∗ ∗) holds for all s ∈ ℂ.
Moreover, the expression for ζ ( s ) is symmetric with respect to s ↦ 1 − s, so ξ ( s ) = ξ ( 1 − s ), s ∈ ℂ. □
nothing: True

html: Corollary 3.5 (Zeroes and poles of z e t a). The ζ function extends to a meromorphic function in ℂ and it has
- (i)
Only one pole, which is a simple pole at s = 1, residue 1.
- (ii)
Simple zeroes at s = − 2 , − 4 , − 6 , ….
- (iii)
Any other zeroes satisfy 0 ≤ Re ⁡ ( s ) ≤ 1.
proof: Proof.
- (ii) – (iii) Follows from the lemma on polynomial growth of ζ on vertical lines.
- (ii) – (iii) We know ζ ( s ) ≠ 0 for Re ⁡ ( s ) > 1. We want to show that if ζ ( s ) = 0 and Re ⁡ ( s ) > 0 then s ∈ { − 2 , − 4 , − 6 , … } and s is a simple zero.
Let Re ⁡ ( s ) > 0. By the functional equation for ζ,
ζ ( s ) = π s − 1 2 ⏟ ≠ 0 Γ ( 1 − s 2 ) Γ ( s 2 ) ζ ( 1 − s ) .
We claim that Γ ( s ) ≠ 0 for all s ∈ ℂ. By the Euler reflection formula,
Γ ( s ) Γ ( 1 − s ) sin ⁡ ( π s ) = π ,
for s ∈ ℂ.
If Γ ( s ) = 0, then Γ has a pole at 1 − s. Hence, 1 − s = − n for some n ≥ 0 integer. But then s = n + 1, and Γ ( n + 1 ) = n ! ≠ 0.
We conclude that for Re ⁡ ( s ) < 0,
ζ ( s ) = 0 ⟺ s 2 is a pole of Γ ⟺ s = − 2 n , n ∈ ℕ
Since the poles of Γ are simple, ζ has a simple zero at s = − 2 n, n ∈ ℕ. □
nothing: True

html: Lemma 3.6 (Borel-Caratheodory Theorem). Assuming that:
- 0 < r < R
- f analytic in B ( 0 , R ) ¯, with f ( 0 ) = 0
Then
sup ⁡ | z | ≤ r | f ( z ) | ≤ 2 r R − r sup ⁡ | z | = R Re ⁡ ( f ( z ) ) .
proof: Proof. This is Exercise 10 on Example Sheet 2. □
nothing: True

html: Lemma 3.7 (Landau). Assuming that:
- z 0 ∈ ℂ and r > 0
- f analytic in B ( z 0 , r )
- for some M > 1 we have | f ( z ) | < e M | f ( z 0 ) | for all z ∈ B ( z 0 , r )
Then for z ∈ B ( z 0 , r ∕ 4 ),
| f ′ ( z ) f ( z ) − ∑ ρ ∈ Z 1 z − ρ | ≤ 9 6 M r ,
where Z is the set of zeroes of f in B ( z 0 , r ∕ 2 ) ¯, counted with multiplicities.
proof: Proof. Let
g ( z ) = f ( z ) ∏ ρ ∈ Z ( z − ρ ) .
Then g is analytic and non-vanishing in B ( z 0 , r ∕ 2 ) ¯. Note that
g ′ ( z ) g ( z ) = f ′ ( z ) f ( z ) − ∑ ρ ∈ Z 1 z − ρ
(( f 1 − f n ) ′ f 1 − f n = ∑ ⁡ i = 1 n f 1 ′ f 1).
Hence, it suffices to prove
| g ′ ( z ) g ( z ) | ≤ 9 6 M r
for z ∈ B ( z 0 , r ∕ 4 ). Write
h ( z ) = g ( z 0 + z ) g ( z 0 ) .
Then h is analytic and non-vanishing in B ( 0 , r ∕ 2 ) ¯, and h ( 0 ) = 1. We want to show
| h ′ ( z ) h ( z ) | ≤ 9 6 M r
for z ∈ B ( 0 , r ∕ 4 ).
For all z 0 ∈ ∂ B ( 0 , r ) we have
| h ( z ) | = | f ( z 0 + z ) f ( z 0 ) ∏ ρ ∈ Z z 0 − ρ z 0 + z − ρ | ≤ | f ( z 0 + z ) f ( z 0 ) | < e M
since
| z 0 − ρ | ≤ r 2 = r − r 2 ≤ | z 0 + z − ρ |
for z ∈ ∂ B ( 0 , r ).
By the maximum modulus principle, | h ( z ) | < e M for z ∈ B ( 0 , r ∕ 2 ) ¯, so
Re ⁡ log ⁡ ⁡ h ( z ) = log ⁡ ⁡ | h ( z ) | < M .
By the Borel-Caratheodory Theorem with radii 3 r 8, r 4 we have for for z ∈ B ( 0 , 3 r ∕ 8 ),
| log ⁡ h ( z ) | ≤ 2 r 4 3 r 8 − r 4 M = 4 M .
Now, for z ∈ B ( 0 , r ∕ 4 ), Cauchy’s theorem gives us
| h ′ ( z ) h ( z ) | = | 1 2 π i ∫ ∂ B ( 0 , 3 r ∕ 8 ) log ⁡ h ( w ) ( z − w ) 2 d w | ≤ 1 2 π ⋅ 2 π 3 r 8 4 M ( 3 r 8 − r 4 ) − 2 = 9 6 M r □
nothing: True

html: Theorem 3.8 (Partial Fraction approximation of z e t a ′ ∕ z e t a).
- (i) Let s = σ + i t with | σ | ≤ 1 0, s ≠ 1 and ζ ( s ) ≠ 0. Then
ζ ′ ( s ) ζ ( s ) = − 1 s − 1 + ∑ | ρ − s | ≤ 1 1 0 1 s − ρ + O ( log ⁡ ( | t | + 2 ) ) .
where the sum is over the zeroes ρ of ζ counted with multiplicity.
- (ii) For any T ≥ 0, there are ≪ log ⁡ ( T + 2 ) many zeroes ρ of ζ (counted with multiplicity) with | Im ⁡ ( ρ ) | ∈ [ T , T + 1 ].
proof: Proof. We apply Landau with z 0 = 2 + i t, r = 5 0, with f ( s ) = ( s − 1 ) ζ ( s ).
By the lemma on polynomial growth of ζ on vertical lines, for σ + i t ∈ B ( z 0 , 5 0 ), we have
| f ( s ) | ≤ C ( | t | + 5 2 ) ( | t + 2 ) 5 0 ≤ C exp ⁡ ( 5 1 log ⁡ ⁡ ( | t | + 5 2 ) ) ≪ C exp ⁡ ( 5 1 log ⁡ ⁡ ( | t | + 5 2 ) ) | f ( z ) |
(| ζ ( z + i t ) | ≍ 1).
Let s ∈ B ( z 0 , 2 5 2 ). Then Landau gives us
f ′ ( s ) f ( s ) = 1 s − 1 + ζ ′ ( s ) ζ ( s ) = ∑ | ρ − z 0 | ≤ 2 S 1 s − ρ + O ( log ⁡ ( | t | + 2 ) ) ( ∗ )
Since B ( z 0 , 2 5 2 ) contains all the points s = σ + i t with | σ | ≤ 1 0, it suffices to show
∑ | ρ − z 0 | ≤ 2 5 | ρ − s | > 1 1 0 1 s − ρ = O ( log ⁡ ( | t | + 2 ) ) . (∗∗)
Substituting s = z 0 in (∗), we get
∑ | ρ − z 0 | ≤ 2 5 1 z 0 − ρ = O ( | ζ ′ ( z 0 ) ζ ( z 0 ) | + 1 + log ⁡ ( | t | + 2 ) ) = O ( log ⁡ ( | t | + 2 ) )
since
| ζ ′ ( 2 + i t ) ζ ( 2 + i t ) | = | ∑ n = 1 ∞ Λ ( n ) n − 2 + i t | ≤ ∑ n = 1 ∞ Λ ( n ) n − 2 = − ζ ( 2 ) ζ ( 2 )
Taking real parts,
∑ | ρ − z 0 | ≤ 2 5 2 − Re ⁡ ( ρ ) | z 0 − ρ | 2 = O ( log ⁡ ( | t | + 2 ) ) .
Since Re ⁡ ( ρ ) ≤ 1,
∑ | ρ − z 0 | ≤ 2 5 1 | z 0 − ρ | 2 = O ( log ⁡ ( | t | + 2 ) ) .
This proves part (ii). It gives also (∗ ∗) since the sum there contains O ( log ⁡ ( | t | + 2 ) ) zeros. □
nothing: True

html: Proposition 3.9. Assuming that:
- σ > 1 and t ∈ ℝ
Then
3 ζ ′ ζ ( σ ) + 4 Re ⁡ ( ζ ′ ζ ( σ + i t ) ) + Re ⁡ ( ζ ′ ζ ( σ + 2 i t ) ) ≤ 0 . (∗∗)
proof: Proof. Recall that
ζ ′ ζ ( s ) = − ∑ n = 1 ∞ Λ ( n ) n s ,
where Λ is the von Mangoldt function function, and Re ⁡ ( s ) > 1.
Taking linear combinations, the LHS of (∗ ∗) becomes
− ∑ n = 1 ∞ Λ ( n ) 3 + 4 Re ⁡ ( n i t ) + Re ⁡ ( n − 2 i t ) n σ = − ∑ n = 1 ∞ Λ ( n ) 3 + 4 cos ⁡ ( t log ⁡ ⁡ n ) + cos ⁡ ( 2 + log ⁡ ⁡ n ) n σ
(Re ⁡ ( n i u ) = cos ⁡ ( u log ⁡ ⁡ n )).
We are done by the inequality:
3 + 4 cos ⁡ α + cos ⁡ 2 α = 2 ( 1 + cos ⁡ α ) 2 ≥ 0
for α ∈ ℝ. □
cloze1: True

html: Theorem 3.10 (Zero-free region). There is a constant c > 0 such that ζ ( σ + i t ) ≠ 0 whenever σ > 1 − c log ⁡ ( | t | + 2 ). In particular, ζ ( s ) ≠ 0 for Re ⁡ ( s ) = 1.
proof: Proof. Let σ ∈ [ 1 , 2 ] and t ∈ ℝ. Suppose ζ ( β + i t ). uwe know that β ≤ 1. We know that ζ has no zeroes in some ball B ( 1 , r ) for some r > 0 (otherwise the entire function ( s − 1 ) ζ ( s ) would have an accumulation point for its zeros).
Choosing c > 0 small enough, we can assume that | t | ≥ r. By the key inequality for ζ ′ ζ,
3 ζ ′ ζ ( σ ) + 4 Re ⁡ ( ζ ′ ζ ( σ + i t ) ) . + Re ⁡ ( ζ ′ ζ ( σ + 2 i t ) ) . ≤ 0 .
Apply partial fraction decomposition of ζ ′ ζ. So
ζ ′ ζ ( s ) = − 1 s − 1 + ∑ | s − ρ | ≤ 1 1 0 1 s − ρ + O ( log ⁡ ( | t | + 2 ) ) . (∗∗)
(t = Im ⁡ ( s )).
Since Re ⁡ ( ρ ) ≤ 1 for any zero ρ,
Re ⁡ 1 σ + i u − ρ = σ − Re ⁡ ( ρ ) | σ + i u − ρ | 2 ≥ 0
(σ > 1).
Discarding terms, we get
− 3 σ − 1 + 4 σ − β ≤ C log ⁡ ( | t | + 2 ) .
Take σ = 1 + s c log ⁡ ( | t | + 2 ), and assume β ≥ 1 − c log ⁡ ( | t | + 2 ) to get
− 3 log ⁡ ( | t | + 2 ) 5 c + 4 log ⁡ ( | t | + 2 ) 6 c ≤ C ( log ⁡ | t | + 2 ) .
Take c = 1 1 6 C to get a contradiction. □
nothing: True

html: Theorem 3.11 (Bounding z e t a ′ ∕ z e t a). Assuming that:
- c > 0 sufficiently small
- T ≥ 0
- Re ⁡ ( s ) ≥ − 1 0
- | Im ⁡ ( s ) | ∈ [ T , T + 1 ]
- s is at least distance c log ⁡ ( T + 2 ) away from any zero or pole
Then
| ζ ′ ( s ) ζ ( s ) | ≪ c ( log ⁡ ( T + 2 ) ) 2 .
proof: Proof. If s = σ + i t with σ > 1 0, then
| ζ ′ ( s ) ζ ( s ) | = | ∑ n = 1 ∞ Λ ( n ) n − s | ≤ ∑ n = 1 ∞ Λ ( n ) n − σ ≪ 1
Assume then that Re ⁡ ( s ) ∈ [ − 1 0 , 1 0 ]. Apply (∗ ∗). Each term satisfies
1 | s − ρ | ≤ log ⁡ ( T + 2 ) c 1 | s − 1 | ≤ log ⁡ ( T + 2 ) c
We know that there are O ( log ⁡ ( T + 2 ) ) zeros with multiplicity having imaginary part ∈ [ T − 2 , T + 2 ]. The claim follows from triangle inequality. □
nothing: True